Problem 22 - Calculus Early Transcendentals

Question

Compute the Jacobian \(J(u, v)\) for the following transformations. $$T: x=u / v, y=v$$

Step-by-Step Solution

Answer

Question: Compute the Jacobian of the transformation with \(x(u,v) = \frac{u}{v}\) and \(y(u,v) = v\). Answer: The Jacobian determinant of the given transformation is \(J(u,v) = \frac{1}{v}\).

Step 1: Compute the partial derivatives

First, we need to find the partial derivatives of the functions \(x(u,v)\) and \(y(u,v)\) with respect to \(u\) and \(v\). Let's find them one by one: Partial derivative of \(x(u,v)\) with respect to \(u\): $$\frac{\partial x}{\partial u} = \frac{\partial (\frac{u}{v})}{\partial u} = \frac{1}{v}$$ Partial derivative of \(x(u,v)\) with respect to \(v\): $$\frac{\partial x}{\partial v} = \frac{\partial (\frac{u}{v})}{\partial v} = -\frac{u}{v^2}$$ Partial derivative of \(y(u,v)\) with respect to \(u\): $$\frac{\partial y}{\partial u} = \frac{\partial v}{\partial u} = 0$$ Partial derivative of \(y(u,v)\) with respect to \(v\): $$\frac{\partial y}{\partial v} = \frac{\partial v}{\partial v} = 1$$ So, the partial derivatives are: $$\frac{\partial x}{\partial u} = \frac{1}{v}, \quad \frac{\partial x}{\partial v} = -\frac{u}{v^2}, \quad \frac{\partial y}{\partial u} = 0, \quad \frac{\partial y}{\partial v} = 1$$

Step 2: Build the Jacobian matrix

Now that we have the required partial derivatives, we can form the Jacobian matrix. The Jacobian matrix of the transformation is given by: $$ J(u,v) = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} $$ Substitute in the computed partial derivatives: $$ J(u,v) = \begin{bmatrix} \frac{1}{v} & -\frac{u}{v^2} \\ 0 & 1 \end{bmatrix} $$

Step 3: Compute the determinant

Finally, we compute the determinant of the Jacobian matrix. The determinant of a 2x2 matrix: $$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ is given by \(ad - bc\). In our case: $$ \det(J(u,v)) = \left(\frac{1}{v}\right)(1) - \left(-\frac{u}{v^2}\right)(0) = \frac{1}{v} $$ So, the Jacobian determinant of the transformation \(T\) is: $$ J(u, v) = \det(J(u,v)) = \frac{1}{v} $$